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Math Help - Multiplicity of a root

  1. #1
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    Multiplicity of a root

    Let c be a fixed real number.
    Show that a root of the equation
    x(x+1)(x+2)...(x+2009)=c
    can have multiplicity at most 2. Determine the number of values of c for which the equation has a root of multiplicity 2.

    Can anybody tell me how to compute this?
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  2. #2
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    Quote Originally Posted by nahduma View Post
    Let c be a fixed real number.
    Show that a root of the equation
    x(x+1)(x+2)...(x+2009)=c
    can have multiplicity at most 2. Determine the number of values of c for which the equation has a root of multiplicity 2.
    If an equation f(x) = c has a root x=a of multiplicity 2 then f'(a) = 0 (I know this is the pre-calculus forum, but I don't think you'll be able to do this question without using differentiation). If the root has multiplicity greater than 2, then also f"(a) = 0.

    For the equation x(x+1)(x+2)...(x+2009) = c, take logs to get \sum_{n=0}^{2009}\ln(x+n) = \ln c. Then let f(x) = \sum_{n=0}^{2009}\ln(x+n). Differentiate to get f'(x) = \sum_{n=0}^{2009}\frac1{x+n}. Then differentiate again to get f''(x) = -\sum_{n=0}^{2009}\frac1{(x+n)^2}. Since all the terms in that last sum have the same sign, it's obvious that their sum cannot be zero. So f(x) = \ln c has no roots of multiplicity greater than 2.

    To find the number of roots of multiplicity 2, notice that the derivative of a function has a root between each pair of consecutive roots of the function. Since x(x+1)(x+2)...(x+2009) = 0 has 2010 roots, its derivative will have 2009 roots.
    Last edited by Opalg; May 2nd 2010 at 01:33 AM.
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